Answer:
0.3726 is the required probability.
Step-by-step explanation:
We are given the following information:
We treat blood sample testing positive for the virus as a success.
P( blood sample testing positive for the virus ) = 0.11
Then the number of blood sample follows a binomial distribution, where
[tex]P(X=x) = \binom{n}{x}.p^x.(1-p)^{n-x}[/tex]
where n is the total number of observations, x is the number of success, p is the probability of success.
Now, we are given n = 4
We have to evaluate the probability of a positive result for four samples combined into one mixture mean that atleast one of the blood sample must be positive for virus.
[tex]P(x \leq 1)\\1 - P(x = 0)\\= 1 - \binom{4}{0}(0.11)^0(1-0.11)^4\\=1-0.6274\\= 0.3726[/tex]
0.3726 is the probability of a positive result for four samples combined into one mixture.